For H2(g) + Br2(g) -> 2HBr(g) k=64
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Ice table:
H2 Br2 HBr
I 0.10 0.10 0
C -x -x +2x
E
Solve for HBr at equilibrium
I do not know how to start this.
I tryed and got .20. Is this correct?
Where did the 1.0, 10.0 and 10.0 come from?
Those were the values that the teacher put in there. I just assumed that since H2 and Br2 are -x and HBr is +2x that HBr would be 2 X 0.10..
Did the teacher put them in as initial concentrations or are they changes or equilibrium concentrations?
He didn't say, he just posted the question like above.
What are you supposed to do with the numbers? What's the question?
To solve for the concentration of HBr at equilibrium, we can use the equilibrium constant expression (K) and the given initial concentrations of H2 and Br2.
Let's start by analyzing the balanced chemical equation: H2(g) + Br2(g) -> 2HBr(g)
The equilibrium constant expression for this reaction is: K = [HBr]^2 / ([H2] * [Br2])
Given K = 64, we can set up the equation: 64 = [HBr]^2 / ([H2] * [Br2])
Now, let's analyze the given initial concentrations in the ice table:
Initial concentrations:
[H2] = 0.10 M
[Br2] = 0.10 M
[HBr] = 0 M (at equilibrium, there is no product yet)
In the ice table:
H2(g) + Br2(g) -> 2HBr(g)
I 0.10 0.10 0
C -x -x +2x
E
Here, 'x' represents the change in molar concentration at equilibrium. Since Stoichiometry shows that for every 1 mole of HBr formed, 1 mole of H2 and 1 mole of Br2 are consumed, the concentration of HBr at equilibrium is '2x'.
Substituting the values from the ice table into the equilibrium constant expression, we have:
64 = (2x)^2 / (0.10 - x) * (0.10 - x)
Simplifying further:
64 = 4x^2 / (0.10 - x)^2
Now, we can solve this equation algebraically for the value of 'x' which will give us the concentration of HBr at equilibrium. However, for simplicity, we will use an approximation technique called the quadratic equation shortcut.
First, let's rewrite the equation:
64 * (0.10 - x)^2 = 4x^2
Expanding and rearranging terms:
6.4 - 12.8x + 6.4x^2 = 4x^2
Simplifying:
6.4x^2 - 12.8x + 6.4 = 4x^2
Combining like terms:
2.4x^2 - 12.8x + 6.4 = 0
At this point, we can use the quadratic equation (x = (-b ± sqrt(b^2 - 4ac)) / 2a) to find the value of 'x'. However, this equation can be simplified by factoring:
2.4x^2 - 12.8x + 6.4 = 0
Factor out common factor:
2.4(x^2 - 5.3x + 2.67) = 0
The equation now becomes:
(x^2 - 5.3x + 2.67) = 0
You can solve this quadratic equation for 'x' using factoring, completing the square, or using the quadratic formula. Once you find the value(s) of 'x', you can substitute it back into the ice table equation to find the concentration of HBr at equilibrium (2x).